The stretch of a spring in millimeters is proportional to the weight in grams attached to the end of the spring. Find the unit rate.

Question

coordinates are (2,10) (3, 15) and (6, 30)

Answer 1

To find the unit rate of the stretch of the spring with respect to the weight attached, we can use the given coordinates that represent weight (in grams) and the corresponding stretch (in millimeters).

The points (x, y) represent the weight (x) and the stretch (y) in the following way:

• (2, 10): 2 grams stretches the spring 10 mm
• (3, 15): 3 grams stretches the spring 15 mm
• (6, 30): 6 grams stretches the spring 30 mm

To find the unit rate, we can calculate the stretch per gram (y/x) for each of these points:

1. For (2, 10): \[ \text{Unit rate} = \frac{y}{x} = \frac{10}{2} = 5 \text{ mm/gram} \]

2. For (3, 15): \[ \text{Unit rate} = \frac{y}{x} = \frac{15}{3} = 5 \text{ mm/gram} \]

3. For (6, 30): \[ \text{Unit rate} = \frac{y}{x} = \frac{30}{6} = 5 \text{ mm/gram} \]

All points yield the same unit rate of 5 mm/gram.

Thus, the unit rate of the stretch of the spring is 5 mm/gram.

Answer 2

Using the points shown in the graph, identify the constant of proportionality for this graph.
k=__
coordinates are (2, 14) (3, 21) and (4, 28)

Answer 3

To find the constant of proportionality \( k \) for the given coordinates, we can use the formula for the constant of proportionality, which is the ratio of \( y \) (the dependent variable) to \( x \) (the independent variable) for each point.

The points given are:

• (2, 14)
• (3, 21)
• (4, 28)

Let's calculate the constant of proportionality \( k \) for each point:

1. For the point (2, 14): \[ k = \frac{y}{x} = \frac{14}{2} = 7 \]

2. For the point (3, 21): \[ k = \frac{y}{x} = \frac{21}{3} = 7 \]

3. For the point (4, 28): \[ k = \frac{y}{x} = \frac{28}{4} = 7 \]

Since all points yield the same value for \( k \), we can conclude that the constant of proportionality for this graph is

\[ k = 7. \]